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Law of total cumulance

In probability theory and mathematical statistics, the law of total cumulance is a generalization to cumulants of the law of total probability, the law of total expectation, and the law of total variance. It has applications in the analysis of time series. It was introduced by David Brillinger.

It is most transparent when stated in its most general form, for joint cumulants, rather than for cumulants of a specified order for just one random variable. In general, we have

$\kappa(X_1,\dots,X_n)=\sum_\pi \kappa(\kappa(X_{\pi_1}\mid Y),\dots,\kappa(X_{\pi_b}\mid Y))\,$

where

the sum is over all partitions $π$ of the set { 1, ..., n } of indices, and

π₁, ..., π_b are all of the "blocks" of the partition π; the expression κ(X_{π_k}) indicates that the joint cumulant of the random variables whose indices are in that block of the partition.

Examples

Only in case n = 2 or 3 is the nth cumulant the same as the nth central moment. The case n = 2 is well-known (see law of total variance). Below is the case n = 3. The notation μ₃ means the third central moment.

$\mu_3(X)=E(\mu_3(X\mid Y))+\mu_3(E(X\mid Y)) +3\,\operatorname{cov}(E(X\mid Y),\operatorname{var}(X\mid Y)).\,$

For general 4th-order cumulants, the rule is as follows:

$\kappa(X_1,X_2,X_3,X_4)\,$

$=\kappa(\kappa(X_1,X_2,X_3,X_4\mid Y))\,$

$\left.\begin{matrix} & +\kappa(\kappa(X_1,X_2,X_3\mid Y),\kappa(X_4\mid Y)) \\ \\ & +\kappa(\kappa(X_1,X_2,X_4\mid Y),\kappa(X_3\mid Y)) \\ \\ & +\kappa(\kappa(X_1,X_3,X_4\mid Y),\kappa(X_2\mid Y)) \\ \\ & +\kappa(\kappa(X_2,X_3,X_4\mid Y),\kappa(X_1\mid Y)) \end{matrix}\right\}(\mathrm{partitions}\ \mathrm{of}\ \mathrm{the}\ 3+1\ \mathrm{form})$

$\left.\begin{matrix} & +\kappa(\kappa(X_1,X_2\mid Y),\kappa(X_3,X_4\mid Y)) \\ \\ & +\kappa(\kappa(X_1,X_3\mid Y),\kappa(X_2,X_4\mid Y)) \\ \\ & +\kappa(\kappa(X_1,X_4\mid Y),\kappa(X_2,X_3\mid Y))\end{matrix}\right\}(\mathrm{partitions}\ \mathrm{of}\ \mathrm{the}\ 2+2\ \mathrm{form})$

$\left.\begin{matrix} & +\kappa(\kappa(X_1,X_2\mid Y),\kappa(X_3\mid Y),\kappa(X_4\mid Y)) \\ \\ & +\kappa(\kappa(X_1,X_3\mid Y),\kappa(X_2\mid Y),\kappa(X_4\mid Y)) \\ \\ & +\kappa(\kappa(X_1,X_4\mid Y),\kappa(X_2\mid Y),\kappa(X_3\mid Y)) \\ \\ & +\kappa(\kappa(X_2,X_3\mid Y),\kappa(X_1\mid Y),\kappa(X_4\mid Y)) \\ \\ & +\kappa(\kappa(X_2,X_4\mid Y),\kappa(X_1\mid Y),\kappa(X_3\mid Y)) \\ \\ & +\kappa(\kappa(X_3,X_4\mid Y),\kappa(X_1\mid Y),\kappa(X_2\mid Y)) \end{matrix}\right\}(\mathrm{partitions}\ \mathrm{of}\ \mathrm{the}\ 2+1+1\ \mathrm{form})$

$+\kappa(\kappa(X_1\mid Y),\kappa(X_2\mid Y),\kappa(X_3\mid Y),\kappa(X_4\mid Y)).\,$

Suppose Y has a Poisson distribution with expected value 1, and X is the sum of Y independent copies of W.

$X=\sum_{y=1}^Y W_y.\,$

Note that all of the cumulants of the Poisson distribution are equal to each other, and so in this case are equal to 1. Also recall that for independent random variables W₁, ..., W_m the nth cumulant is additive:

$\kappa_n(W_1+\cdots+W_m)=\kappa_n(W_1)+\cdots+\kappa_n(W_m).\,$

Suppose we want the 4th cumulant of X. We have:

$\kappa_4(X)=\kappa(X,X,X,X)\,$

$=\kappa_1(\kappa_4(X\mid Y))+4\kappa(\kappa_3(X\mid Y),\kappa_1(X\mid Y))+3\kappa_2(\kappa_2(X\mid Y))\,$

$+6\kappa(\kappa_2(X\mid Y),\kappa_1(X\mid Y),\kappa_1(X\mid Y))+\kappa_4(\kappa_1(X\mid Y))\,$

$=\kappa_1(Y\kappa_4(W))+4\kappa(Y\kappa_3(W),Y\kappa_1(W)) +3\kappa_2(Y\kappa_2(W))\,$

$+6\kappa(Y\kappa_2(W),Y\kappa_1(W),Y\kappa_1(W)) +\kappa_4(Y\kappa_1(W))\,$

$=\kappa_4(W)\kappa_1(Y)+4\kappa_3(W)\kappa_1(W)\kappa_2(Y) +3\kappa_2(W)^2 \kappa_2(Y)\,$

$+6\kappa_2(W) \kappa_1(W)^2 \kappa_3(Y)+\kappa_2(W)^4 \kappa_4(Y).\,$

$=\kappa_4(W)+4\kappa_3(W)\kappa_1(W) +3\kappa_2(W)^2+6\kappa_2(W) \kappa_1(W)^2+\kappa_2(W)^4.\,$

Now we recognize this last sum as the sum over all partitions of the set { 1, 2, 3, 4 }, of the product over all blocks of the partition, of cumulants of W of order equal to the size of the block (see cumulant for a more leisurely discussion). That is precisely the 4th moment of W. Hence the moments of W are the cumulants of X. In this way we see that every moment sequence is also a cumulant sequence (the converse cannot be true, since cumulants of even order ≥ 4 can be negative).

References

David Brillinger, "The calculation of cumulants via conditioning", Annals of the Institute of Statistical Mathematics, Vol. 21 (1969), pp. 215-218.

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10-26-2009 08:16:03
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